Question

A man slides down on a telegraphic pole with an acceleration equal to ${\left(\frac{1}{4}\right)}^{th}$ of acceleration due to gravity. The frictional force between man and pole is equal to in terms of man's weight $w$

• A. $\frac{w}{4}$
• B. $\frac{w}{2}$
• C. $\frac{3w}{4}$
• D. $w$

Answer 1

The correct option is C

$\frac{3w}{4}$

Step 1: Given data:

Acceleration of the man, $a=\frac{g}{4}$

Weight of the man, $w$$=mg$

Step 2: Balancing forces acting on man:

Friction force $f$ will act opposite to the weight of a man, $w$

$ma=w–f$ [$m$ is mass, $g$is acceleration due to gravity]

$\Rightarrow f=w–ma$

$\Rightarrow f=mg–m\times \frac{g}{4}$ $\left[\because a=\frac{g}{4},w=mg\right]$

Therefore, $f$$=\frac{3w}{4}$

Hence, option (C) is correct.